Question

In CaCo3 + HCl = CaCl2 + H2O +CO3, how many liters of carbon dioxide gas, measured at STP, can be obtained from 45.0 g of calcium carbonate?

A) 19.8 L


B) 10.1 L


C) 22.4 L


D)2.28 L

Answer 1

B. 10.1L

Step-by-step explanation:

this is correct just did the worksheet with this on it.

Answer 2

B) 10.1 L

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction which should be corrected as shown below:

`CaCO_3 + HCl \rightarrow CaCl_2 + H_2O +CO_2`

Since 45.0g of calcium carbonate are used, the produced moles of carbon dioxide, via stoichiometry, are found to be:

`n_(CO_2)=45.0g CaCO_3*(1molCaCO_3)/(100gCaCO_3)*(1molCO_2)/(1molCaCO_3)=0.45molCO_2`

Finally, since STP conditions are referred to a temperature of 273.15K and 1 atm, the volume, by using the ideal gas equation result:

`V=(nRT)/(P)=(0.45mol*0.082(atm*L)/(mol*K)*273.15K )/(1atm) \\ \\V=10.1L`

So the answer is B) 10.1 L.

Best regards.